3.17.97 \(\int \frac {\sqrt {1-2 x}}{(2+3 x)^4 (3+5 x)} \, dx\)

Optimal. Leaf size=113 \[ \frac {1207 \sqrt {1-2 x}}{49 (3 x+2)}+\frac {52 \sqrt {1-2 x}}{21 (3 x+2)^2}+\frac {\sqrt {1-2 x}}{3 (3 x+2)^3}+\frac {83264 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}}-50 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {99, 151, 156, 63, 206} \begin {gather*} \frac {1207 \sqrt {1-2 x}}{49 (3 x+2)}+\frac {52 \sqrt {1-2 x}}{21 (3 x+2)^2}+\frac {\sqrt {1-2 x}}{3 (3 x+2)^3}+\frac {83264 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}}-50 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - 2*x]/((2 + 3*x)^4*(3 + 5*x)),x]

[Out]

Sqrt[1 - 2*x]/(3*(2 + 3*x)^3) + (52*Sqrt[1 - 2*x])/(21*(2 + 3*x)^2) + (1207*Sqrt[1 - 2*x])/(49*(2 + 3*x)) + (8
3264*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(49*Sqrt[21]) - 50*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin {align*} \int \frac {\sqrt {1-2 x}}{(2+3 x)^4 (3+5 x)} \, dx &=\frac {\sqrt {1-2 x}}{3 (2+3 x)^3}-\frac {1}{3} \int \frac {-18+25 x}{\sqrt {1-2 x} (2+3 x)^3 (3+5 x)} \, dx\\ &=\frac {\sqrt {1-2 x}}{3 (2+3 x)^3}+\frac {52 \sqrt {1-2 x}}{21 (2+3 x)^2}-\frac {1}{42} \int \frac {-1374+1560 x}{\sqrt {1-2 x} (2+3 x)^2 (3+5 x)} \, dx\\ &=\frac {\sqrt {1-2 x}}{3 (2+3 x)^3}+\frac {52 \sqrt {1-2 x}}{21 (2+3 x)^2}+\frac {1207 \sqrt {1-2 x}}{49 (2+3 x)}-\frac {1}{294} \int \frac {-59124+36210 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx\\ &=\frac {\sqrt {1-2 x}}{3 (2+3 x)^3}+\frac {52 \sqrt {1-2 x}}{21 (2+3 x)^2}+\frac {1207 \sqrt {1-2 x}}{49 (2+3 x)}-\frac {41632}{49} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx+1375 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {\sqrt {1-2 x}}{3 (2+3 x)^3}+\frac {52 \sqrt {1-2 x}}{21 (2+3 x)^2}+\frac {1207 \sqrt {1-2 x}}{49 (2+3 x)}+\frac {41632}{49} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-1375 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {\sqrt {1-2 x}}{3 (2+3 x)^3}+\frac {52 \sqrt {1-2 x}}{21 (2+3 x)^2}+\frac {1207 \sqrt {1-2 x}}{49 (2+3 x)}+\frac {83264 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}}-50 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 83, normalized size = 0.73 \begin {gather*} \frac {\sqrt {1-2 x} \left (10863 x^2+14848 x+5087\right )}{49 (3 x+2)^3}+\frac {83264 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}}-50 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - 2*x]/((2 + 3*x)^4*(3 + 5*x)),x]

[Out]

(Sqrt[1 - 2*x]*(5087 + 14848*x + 10863*x^2))/(49*(2 + 3*x)^3) + (83264*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(49*S
qrt[21]) - 50*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.46, size = 95, normalized size = 0.84 \begin {gather*} -\frac {2 \sqrt {1-2 x} \left (10863 (1-2 x)^2-51422 (1-2 x)+60907\right )}{49 (3 (1-2 x)-7)^3}+\frac {83264 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}}-50 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[1 - 2*x]/((2 + 3*x)^4*(3 + 5*x)),x]

[Out]

(-2*(60907 - 51422*(1 - 2*x) + 10863*(1 - 2*x)^2)*Sqrt[1 - 2*x])/(49*(-7 + 3*(1 - 2*x))^3) + (83264*ArcTanh[Sq
rt[3/7]*Sqrt[1 - 2*x]])/(49*Sqrt[21]) - 50*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

________________________________________________________________________________________

fricas [A]  time = 1.32, size = 130, normalized size = 1.15 \begin {gather*} \frac {25725 \, \sqrt {55} {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 41632 \, \sqrt {21} {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (\frac {3 \, x - \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) + 21 \, {\left (10863 \, x^{2} + 14848 \, x + 5087\right )} \sqrt {-2 \, x + 1}}{1029 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(1/2)/(2+3*x)^4/(3+5*x),x, algorithm="fricas")

[Out]

1/1029*(25725*sqrt(55)*(27*x^3 + 54*x^2 + 36*x + 8)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 41632
*sqrt(21)*(27*x^3 + 54*x^2 + 36*x + 8)*log((3*x - sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) + 21*(10863*x^2 + 14
848*x + 5087)*sqrt(-2*x + 1))/(27*x^3 + 54*x^2 + 36*x + 8)

________________________________________________________________________________________

giac [A]  time = 1.23, size = 123, normalized size = 1.09 \begin {gather*} 25 \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {41632}{1029} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {10863 \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - 51422 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 60907 \, \sqrt {-2 \, x + 1}}{196 \, {\left (3 \, x + 2\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(1/2)/(2+3*x)^4/(3+5*x),x, algorithm="giac")

[Out]

25*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 41632/1029*sqrt(21)*
log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 1/196*(10863*(2*x - 1)^2*sqrt(-2*
x + 1) - 51422*(-2*x + 1)^(3/2) + 60907*sqrt(-2*x + 1))/(3*x + 2)^3

________________________________________________________________________________________

maple [A]  time = 0.01, size = 75, normalized size = 0.66 \begin {gather*} \frac {83264 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{1029}-50 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )-\frac {54 \left (\frac {1207 \left (-2 x +1\right )^{\frac {5}{2}}}{147}-\frac {7346 \left (-2 x +1\right )^{\frac {3}{2}}}{189}+\frac {1243 \sqrt {-2 x +1}}{27}\right )}{\left (-6 x -4\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x+1)^(1/2)/(3*x+2)^4/(5*x+3),x)

[Out]

-50*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)-54*(1207/147*(-2*x+1)^(5/2)-7346/189*(-2*x+1)^(3/2)+1243/27
*(-2*x+1)^(1/2))/(-6*x-4)^3+83264/1029*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

________________________________________________________________________________________

maxima [A]  time = 1.54, size = 128, normalized size = 1.13 \begin {gather*} 25 \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {41632}{1029} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {2 \, {\left (10863 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 51422 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 60907 \, \sqrt {-2 \, x + 1}\right )}}{49 \, {\left (27 \, {\left (2 \, x - 1\right )}^{3} + 189 \, {\left (2 \, x - 1\right )}^{2} + 882 \, x - 98\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(1/2)/(2+3*x)^4/(3+5*x),x, algorithm="maxima")

[Out]

25*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 41632/1029*sqrt(21)*log(-(sqrt
(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 2/49*(10863*(-2*x + 1)^(5/2) - 51422*(-2*x + 1)^(3/2
) + 60907*sqrt(-2*x + 1))/(27*(2*x - 1)^3 + 189*(2*x - 1)^2 + 882*x - 98)

________________________________________________________________________________________

mupad [B]  time = 1.20, size = 89, normalized size = 0.79 \begin {gather*} \frac {83264\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{1029}-50\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )+\frac {\frac {2486\,\sqrt {1-2\,x}}{27}-\frac {14692\,{\left (1-2\,x\right )}^{3/2}}{189}+\frac {2414\,{\left (1-2\,x\right )}^{5/2}}{147}}{\frac {98\,x}{3}+7\,{\left (2\,x-1\right )}^2+{\left (2\,x-1\right )}^3-\frac {98}{27}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - 2*x)^(1/2)/((3*x + 2)^4*(5*x + 3)),x)

[Out]

(83264*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/1029 - 50*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11) +
((2486*(1 - 2*x)^(1/2))/27 - (14692*(1 - 2*x)^(3/2))/189 + (2414*(1 - 2*x)^(5/2))/147)/((98*x)/3 + 7*(2*x - 1
)^2 + (2*x - 1)^3 - 98/27)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)**(1/2)/(2+3*x)**4/(3+5*x),x)

[Out]

Timed out

________________________________________________________________________________________